Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int maxn = 10001, maxs = 1000001;
int n, x[maxn], y[maxn], xsize, xop[maxs][2], ysize, yop[maxs][2];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", x + i);
  for (int i = 0; i < n; ++i) scanf("%d", y + i);
  for (int i = 0; i < 30 && i < n; ++i)
    for (int j = i + 1; j < n; ++j) {
      if (x[i] < x[j]) {
        x[i] ^= x[j];
        xop[xsize][0] = i, xop[xsize++][1] = j;
        x[j] ^= x[i];
        xop[xsize][0] = j, xop[xsize++][1] = i;
        x[i] ^= x[j];
        xop[xsize][0] = i, xop[xsize++][1] = j;
      }
      if ((x[i] ^ x[j]) < x[j]) {
        x[j] ^= x[i];
        xop[xsize][0] = j, xop[xsize++][1] = i;
      }
    }
  for (int i = n - 1 < 29 ? n - 1 : 29; i >= 0; --i)
    for (int j = 0; j < i; ++j)
      if ((x[i] ^ x[j]) < x[j]) {
        x[j] ^= x[i];
        xop[xsize][0] = j, xop[xsize++][1] = i;
      }
  for (int i = 0; i < 30 && i < n; ++i)
    for (int j = i + 1; j < n; ++j) {
      if (y[i] < y[j]) {
        y[i] ^= y[j];
        yop[ysize][0] = i, yop[ysize++][1] = j;
        y[j] ^= y[i];
        yop[ysize][0] = j, yop[ysize++][1] = i;
        y[i] ^= y[j];
        yop[ysize][0] = i, yop[ysize++][1] = j;
      }
      if ((y[i] ^ y[j]) < y[j]) {
        y[j] ^= y[i];
        yop[ysize][0] = j, yop[ysize++][1] = i;
      }
    }
  for (int i = n - 1 < 29 ? n - 1 : 29; i >= 0; --i)
    for (int j = 0; j < i; ++j)
      if ((y[i] ^ y[j]) < y[j]) {
        y[j] ^= y[i];
        yop[ysize][0] = j, yop[ysize++][1] = i;
      }
  for (int i = 0; i < 30 && i < n; ++i) {
    if ((int)log2(x[i]) > (int)log2(y[i])) {
      x[i] ^= x[i];
      xop[xsize][0] = i, xop[xsize++][1] = i;
    }
    for (int j = i + 1; j < 30 && j < n; ++j)
      if ((x[i] ^ x[j]) <= y[i]) {
        x[i] ^= x[j];
        xop[xsize][0] = i, xop[xsize++][1] = j;
      }
    if (x[i] != y[i]) {
      puts("-1");
      return 0;
    }
  }
  printf("%d\n", xsize + ysize);
  for (int i = 0; i < xsize; ++i)
    printf("%d %d\n", xop[i][0] + 1, xop[i][1] + 1);
  for (int i = ysize - 1; i >= 0; --i)
    printf("%d %d\n", yop[i][0] + 1, yop[i][1] + 1);
  return 0;
}